Simplify the following expression: $y = \dfrac{-9x^2+29x- 6}{x - 3}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-9)}{(-6)} &=& 54 \\ {a} + {b} &=& &=& {29} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $54$ and add them together. The factors that add up to ${29}$ will be your ${a}$ and ${b}$ When ${a}$ is ${2}$ and ${b}$ is ${27}$ $ \begin{eqnarray} {ab} &=& ({2})({27}) &=& 54 \\ {a} + {b} &=& {2} + {27} &=& 29 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-9}x^2 +{2}x) + ({27}x {-6}) $ Factor out the common factors: $ x(-9x + 2) - 3(-9x + 2)$ Now factor out $(-9x + 2)$ $ (-9x + 2)(x - 3)$ The original expression can therefore be written: $ \dfrac{(-9x + 2)(x - 3)}{x - 3}$ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ This leaves us with $-9x + 2; x \neq 3$.